Solve (a_{4}+5/2)^2=(a_{4}-a_{1})(a_{4}+7d)

$$( a _ { 4 } + \frac { 5 } { 2 } ) ^ { 2 } = ( a _ { 4 } - a _ { 1 } ) ( a _ { 4 } + 7 d )$$

Solve for a_1

$$\left\{\begin{matrix}a_{1}=\frac{28a_{4}d-20a_{4}-25}{4\left(a_{4}+7d\right)}\text{, }&a_{4}\neq -7d\\a_{1}\in \mathrm{R}\text{, }&a_{4}=-\frac{5}{2}\text{ and }d=\frac{5}{14}\end{matrix}\right.$$

Steps for Solving Linear Equation

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(a_{4}+\frac{5}{2}\right)^{2}.

$$a_{4}^{2}+5a_{4}+\frac{25}{4}=\left(a_{4}-a_{1}\right)\left(a_{4}+7d\right)$$

Use the distributive property to multiply a_{4}-a_{1} by a_{4}+7d.

$$a_{4}^{2}+5a_{4}+\frac{25}{4}=a_{4}^{2}+7a_{4}d-a_{1}a_{4}-7a_{1}d$$

Swap sides so that all variable terms are on the left hand side.

$$a_{4}^{2}+7a_{4}d-a_{1}a_{4}-7a_{1}d=a_{4}^{2}+5a_{4}+\frac{25}{4}$$

Subtract a_{4}^{2} from both sides.

$$7a_{4}d-a_{1}a_{4}-7a_{1}d=a_{4}^{2}+5a_{4}+\frac{25}{4}-a_{4}^{2}$$

Combine a_{4}^{2} and -a_{4}^{2} to get 0.

$$7a_{4}d-a_{1}a_{4}-7a_{1}d=5a_{4}+\frac{25}{4}$$

Subtract 7a_{4}d from both sides.

$$-a_{1}a_{4}-7a_{1}d=5a_{4}+\frac{25}{4}-7a_{4}d$$

Combine all terms containing a_{1}.

$$\left(-a_{4}-7d\right)a_{1}=5a_{4}+\frac{25}{4}-7a_{4}d$$

The equation is in standard form.

$$\left(-a_{4}-7d\right)a_{1}=\frac{25}{4}+5a_{4}-7a_{4}d$$

Divide both sides by -a_{4}-7d.

$$\frac{\left(-a_{4}-7d\right)a_{1}}{-a_{4}-7d}=\frac{\frac{25}{4}+5a_{4}-7a_{4}d}{-a_{4}-7d}$$

Dividing by -a_{4}-7d undoes the multiplication by -a_{4}-7d.

$$a_{1}=\frac{\frac{25}{4}+5a_{4}-7a_{4}d}{-a_{4}-7d}$$

Divide 5a_{4}+\frac{25}{4}-7a_{4}d by -a_{4}-7d.

$$a_{1}=-\frac{25+20a_{4}-28a_{4}d}{4\left(a_{4}+7d\right)}$$

Solve for a_4

$$\left\{\begin{matrix}a_{4}=-\frac{28a_{1}d+25}{4\left(a_{1}-7d+5\right)}\text{, }&a_{1}\neq 7d-5\\a_{4}\in \mathrm{R}\text{, }&d=\frac{5}{14}\text{ and }a_{1}=-\frac{5}{2}\end{matrix}\right.$$

Steps for Solving Linear Equation

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(a_{4}+\frac{5}{2}\right)^{2}.

$$a_{4}^{2}+5a_{4}+\frac{25}{4}=\left(a_{4}-a_{1}\right)\left(a_{4}+7d\right)$$

Use the distributive property to multiply a_{4}-a_{1} by a_{4}+7d.

$$a_{4}^{2}+5a_{4}+\frac{25}{4}=a_{4}^{2}+7a_{4}d-a_{1}a_{4}-7a_{1}d$$

Subtract a_{4}^{2} from both sides.

$$a_{4}^{2}+5a_{4}+\frac{25}{4}-a_{4}^{2}=7a_{4}d-a_{1}a_{4}-7a_{1}d$$

Combine a_{4}^{2} and -a_{4}^{2} to get 0.

$$5a_{4}+\frac{25}{4}=7a_{4}d-a_{1}a_{4}-7a_{1}d$$

Subtract 7a_{4}d from both sides.

$$5a_{4}+\frac{25}{4}-7a_{4}d=-a_{1}a_{4}-7a_{1}d$$

Add a_{1}a_{4} to both sides.

$$5a_{4}+\frac{25}{4}-7a_{4}d+a_{1}a_{4}=-7a_{1}d$$

Subtract \frac{25}{4} from both sides.

$$5a_{4}-7a_{4}d+a_{1}a_{4}=-7a_{1}d-\frac{25}{4}$$

Combine all terms containing a_{4}.

$$\left(5-7d+a_{1}\right)a_{4}=-7a_{1}d-\frac{25}{4}$$

The equation is in standard form.

$$\left(a_{1}-7d+5\right)a_{4}=-7a_{1}d-\frac{25}{4}$$

Divide both sides by -7d+5+a_{1}.

$$\frac{\left(a_{1}-7d+5\right)a_{4}}{a_{1}-7d+5}=\frac{-7a_{1}d-\frac{25}{4}}{a_{1}-7d+5}$$

Dividing by -7d+5+a_{1} undoes the multiplication by -7d+5+a_{1}.

$$a_{4}=\frac{-7a_{1}d-\frac{25}{4}}{a_{1}-7d+5}$$

Divide -7a_{1}d-\frac{25}{4} by -7d+5+a_{1}.

$$a_{4}=-\frac{28a_{1}d+25}{4\left(a_{1}-7d+5\right)}$$